Physical Pendulum: g from T(δ)
A uniform rod of known length. Time the small-angle period at several pivot distances from the centre of mass and recover g from a fit of T² versus (L²/12δ + δ).
Goal
Determine g from T² = (4π²/g)(L²/(12δ) + δ). The slope of T² versus (L²/(12δ)+δ) is 4π²/g.
Equipment
- Uniform rod (known L)
- Movable knife-edge
- Stopwatch
Experiment
Theory
For a uniform rod, I_cm = mL²/12. Parallel-axis: I = I_cm + mδ². The small-angle period is T = 2π√(I/(mgδ)), so T² = (4π²/g)(L²/(12δ)+δ). The bench hides T, I and L_eq. L is known. This is not the simple-pendulum lab (there T² ∝ L).
Procedure
- Rod length L is fixed and known. You only change the pivot distance δ from the centre of mass. Stay away from the CM (δ = 0 is undefined).
- Record. A stopwatch logs the small-angle period T with small timing noise. There is no live T, I or L_eq.
- The notebook computes T² and L²/(12δ)+δ. Repeat for at least 6 values of δ from about 0.12 m to 0.50 m.
- Fit T² versus (L²/(12δ)+δ); g = 4π² / slope. Compare with the reference.
Conclusion
The fitted g agrees with the reference. Main uncertainties: stopwatch noise, the small-angle approximation, and treating the rod as uniform.